3.3.98 \(\int \frac {x^2 (d^2-e^2 x^2)^p}{(d+e x)^4} \, dx\) [298]

3.3.98.1 Optimal result

Integrand size = 25, antiderivative size = 163 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}-\frac {2^{-3+p} (7+p) \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (1+p)} \]

output

-1/2*d*(-e^2*x^2+d^2)^(p+1)/e^3/(3-p)/(e*x+d)^4+(-e^2*x^2+d^2)^(p+1)/e^3/( 
1-2*p)/(e*x+d)^3-2^(-3+p)*(7+p)*(1+e*x/d)^(-1-p)*(-e^2*x^2+d^2)^(p+1)*hype 
rgeom([p+1, 3-p],[2+p],1/2*(-e*x+d)/d)/d^3/e^3/(2*p^3-5*p^2-4*p+3)
 

3.3.98.2 Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {2^{-4+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )-4 \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+\operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{d^2 e^3 (1+p)} \]

input

Integrate[(x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]
 

output

-((2^(-4 + p)*(d - e*x)*(d^2 - e^2*x^2)^p*(4*Hypergeometric2F1[2 - p, 1 + 
p, 2 + p, (d - e*x)/(2*d)] - 4*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - 
 e*x)/(2*d)] + Hypergeometric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/( 
d^2*e^3*(1 + p)*(1 + (e*x)/d)^p))
 

3.3.98.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {581, 25, 27, 671, 473, 79}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx\)

\(\Big \downarrow \) 581

\(\displaystyle \frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}-\frac {\int -\frac {d (3 d+2 e (p+1) x) \left (d^2-e^2 x^2\right )^p}{(d+e x)^4}dx}{e^2 (1-2 p)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {d (3 d+2 e (p+1) x) \left (d^2-e^2 x^2\right )^p}{(d+e x)^4}dx}{e^2 (1-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \int \frac {(3 d+2 e (p+1) x) \left (d^2-e^2 x^2\right )^p}{(d+e x)^4}dx}{e^2 (1-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {d \left (\frac {(p+7) \int \frac {\left (d^2-e^2 x^2\right )^p}{(d+e x)^3}dx}{3-p}-\frac {(1-2 p) \left (d^2-e^2 x^2\right )^{p+1}}{2 e (3-p) (d+e x)^4}\right )}{e^2 (1-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {d \left (\frac {(p+7) (d-e x)^{-p-1} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \int (d-e x)^p \left (\frac {e x}{d}+1\right )^{p-3}dx}{d^4 (3-p)}-\frac {(1-2 p) \left (d^2-e^2 x^2\right )^{p+1}}{2 e (3-p) (d+e x)^4}\right )}{e^2 (1-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}+\frac {d \left (-\frac {(1-2 p) \left (d^2-e^2 x^2\right )^{p+1}}{2 e (3-p) (d+e x)^4}-\frac {2^{p-3} (p+7) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (3-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d^4 e (3-p) (p+1)}\right )}{e^2 (1-2 p)}\)

input

Int[(x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]
 

output

(d^2 - e^2*x^2)^(1 + p)/(e^3*(1 - 2*p)*(d + e*x)^3) + (d*(-1/2*((1 - 2*p)* 
(d^2 - e^2*x^2)^(1 + p))/(e*(3 - p)*(d + e*x)^4) - (2^(-3 + p)*(7 + p)*(1 
+ (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1 + p)*Hypergeometric2F1[3 - p, 1 + p 
, 2 + p, (d - e*x)/(2*d)])/(d^4*e*(3 - p)*(1 + p))))/(e^2*(1 - 2*p))
 

3.3.98.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 581

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b*x^ 
2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m 
+ c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p 
)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & 
& IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] 
)
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

3.3.98.4 Maple [F]

\[\int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}d x\]

input

int(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x)
 

output

int(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x)
 

3.3.98.5 Fricas [F]

\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]

input

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="fricas")
 

output

integral((-e^2*x^2 + d^2)^p*x^2/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4 
*d^3*e*x + d^4), x)
 

3.3.98.6 Sympy [F]

\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \]

input

integrate(x**2*(-e**2*x**2+d**2)**p/(e*x+d)**4,x)
 

output

Integral(x**2*(-(-d + e*x)*(d + e*x))**p/(d + e*x)**4, x)
 

3.3.98.7 Maxima [F]

\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]

input

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="maxima")
 

output

integrate((-e^2*x^2 + d^2)^p*x^2/(e*x + d)^4, x)
 

3.3.98.8 Giac [F]

\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]

input

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="giac")
 

output

integrate((-e^2*x^2 + d^2)^p*x^2/(e*x + d)^4, x)
 

3.3.98.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \]

input

int((x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x)
 

output

int((x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4, x)